You have a standard pack of 52 playing cards. You then shuffle them and begin to draw out cards until you have three of a kind. What is the most likely number of cards drawn when this happens?

You then shuffle another pack of 52 playing cards into the pile. What happens to the expected number of cards now? (i.e. does it double / halve / stay the same?)

(In reply to re: A non-poker solution (computer program not used) by Charlie)

By the way, the remaining figures on your table can't even be the cumulative version of the numbers you calculate, that is, of the prob of completing an ace triplet by that number. That comes out to the last column of the following table:
3 0.000180995 0.000180995
4 0.000531905 0.000712901
5 0.001041647 0.001754548
6 0.001699141 0.003453689
7 0.002493305 0.005946994
8 0.003413058 0.009360052
9 0.004447317 0.013807369
10 0.005585003 0.019392372
11 0.006815034 0.026207406
12 0.008126327 0.034333733
13 0.009507803 0.043841537
14 0.010948379 0.054789916
15 0.012436975 0.067226891
16 0.013962508 0.081189399
17 0.015513898 0.096703297
18 0.017080063 0.113783359
19 0.018649922 0.132433281
20 0.020212393 0.152645674
21 0.021756395 0.174402069
22 0.023270847 0.197672915
23 0.024744667 0.222417582
24 0.026166774 0.248584357
25 0.027526087 0.276110444
26 0.028811525 0.304921969
27 0.030012005 0.334933974
28 0.031116447 0.366050420
29 0.032113769 0.398164189
30 0.032992889 0.431157078
31 0.033742728 0.464899806
32 0.034352202 0.499252008
33 0.034810232 0.534062240
34 0.035105735 0.569167975
35 0.035227630 0.604395604
36 0.035164835 0.639560440
37 0.034906270 0.674466710
38 0.034440853 0.708907563
39 0.033757503 0.742665066
40 0.032845138 0.775510204

And certainly there's no hope of multiplying these by 13 and getting anything that resembles a probability.

Posted by Charlie on 2003-11-20 22:43:55